Question: Vicente has a prism-like water tank whose base area is $1.2$ square meters. He bought $6$ goldfish at the store, and the store owner told him to make sure their density in the tank isn't more than $4$ fish per cubic meter. Vicente needs to figure out how high to fill the water in the tank. What is the lowest possible height so the fish aren't too crowded?
Explanation: This is a density word problem. To solve it, we can use the following equation, which is the volume definition of density: ${\text{Density}}=\dfrac{{\text{Total quantity}}}{{\text{Volume}}}$ What do we know? The base area of the prism-like tank is $1.2$ square meters (we can use this to find the ${\text{volume}}$ ). The number of fish, which is the ${\text{total quantity}}$, is $ 6$. The ${\text{density}}$ of the fish shouldn't be more than $ 4$ fish per cubic meter. What do we need to find? The height of the water, which together with the base area of the tank gives us the ${\text{volume}}$ of the water. We want the ${\text{density}}$ to be ${4}$ fish per cubic meter at most. So this is the inequality we want to solve: $\begin{aligned} {\text{Density}} &\leq {4}\\\\ \dfrac{{\text{Total quantity}}}{{\text{Volume}}} &\leq {4} \end{aligned}$ Let's denote the height of the water as $h$. Then, the ${\text{volume}}$ of the water is ${1.2h}$ cubic meters. Now we can plug ${\text{total quantity}=6}$ and ${\text{volume}=1.2h}$ in the inequality. $\begin{aligned} \dfrac{{\text{Total quantity}}}{{\text{Volume}}} &\leq {4} \\\\ \dfrac{{6}}{{1.2h}} &\leq {4} \\\\ \cancel{1.2h}\cdot\dfrac{6}{\cancel{1.2h}} &\leq 1.2h\cdot 4 \\\\ 6 &\leq 4.8h \\\\ 1.25 &\leq h \end{aligned}$ The height must be $1.25$ meters or more. If the height is less, the density will be more than $4$, which is too high. Therefore, the lowest possible height so the fish aren't too crowded is $1.25$ meters.